2.7: Linearization and Newton’s Method

Learning Objectives

A student will be able to:

Approximate a function by the method of linearization.

Know Newton’s Method for approximating roots of a function.

Linearization: The Tangent Line Approximation

If \begin{align*}f\end{align*} is a differentiable function at \begin{align*}x_0\end{align*}, then the tangent line, \begin{align*}y = mx + b\end{align*}, to the curve \begin{align*}y = f(x)\end{align*} at \begin{align*}x_0\end{align*} is a good approximation to the curve \begin{align*}y = f(x)\end{align*} for values of \begin{align*}x\end{align*} near \begin{align*}x_0\end{align*} (Figure 8a). If you “zoom in” on the two graphs, \begin{align*}y = f(x)\end{align*} and the tangent line, at the point of tangency, \begin{align*}(x_0, f(x_0))\end{align*}, or if you look at a table of values near the point of tangency, you will notice that the values are very close (Figure 8b).

Since the tangent line passes through point \begin{align*}(x_0, f(x_0))\end{align*} and the slope is \begin{align*}f'(x_0)\end{align*}, we can write the equation of the tangent line, in point-slope form, as

So for values of \begin{align*}x\end{align*} close to \begin{align*}x_0\end{align*}, the values of \begin{align*}y\end{align*} of this tangent line will closely approximate \begin{align*}f(x)\end{align*}. This gives the approximation

\begin{align*}f(x) = f(x_0) + f'(x_0) (x - x_0).\end{align*}

The Tangent Line Approximation (Linearization)

If \begin{align*}f\end{align*} is a differentiable function at \begin{align*}x = x_0\end{align*}, then the approximation function

This tells us that near the point \begin{align*}x = 1\end{align*}, the function \begin{align*} f(x) = \sqrt{x + 3}\end{align*} approximates the line \begin{align*}y = (x/4) + 7/4\end{align*}. As we move away from \begin{align*}x = 1\end{align*}, we lose accuracy (Figure 9).

Newton’s Method

When faced with a mathematical problem that cannot be solved with simple algebraic means, such as finding the roots of the polynomial \begin{align*}x^3 -2x + 3 = 0,\end{align*} calculus sometimes provides a way of finding the approximate solutions.

Let's say we are interested in computing \begin{align*} \sqrt{5}\end{align*} without using a calculator or a table. To do so, think about this problem in a different way. Assume that we are interested in solving the quadratic equation

\begin{align*}f(x) = x^2 - 5 = 0\end{align*}

which leads to the roots \begin{align*} x = \pm \sqrt{5}\end{align*}.

The idea here is to find the linearization of the above function, which is a straight-line equation, and then solve the linear equation for \begin{align*}x\end{align*}.

Since

\begin{align*} \sqrt{4} < \sqrt{5} < \sqrt{9}\end{align*}

or

\begin{align*} 2 < \sqrt{5} < 3,\end{align*}

We choose the linear approximation of \begin{align*}f(x)\end{align*} to be near \begin{align*}x_0 = 2\end{align*}. Since \begin{align*}f(x) = x^2 - 5,\end{align*}\begin{align*}f'(x) = 2x\end{align*} and thus \begin{align*}f(2) = -1\end{align*} and \begin{align*}f'(2) = 4.\end{align*} Using the linear approximation formula,

If you use a calculator, you will get \begin{align*}x = 2.236 \ldots\end{align*} As you can see, this is a fairly good approximation. To be sure, calculate the percent difference\begin{align*}[\%\;\mathrm{diff}]\end{align*} between the actual value and the approximate value,

We can actually make our approximation even better by repeating what we have just done not for \begin{align*}x = 2\end{align*}, but for \begin{align*}x_1 = 2.25 = \frac{9} {4}\end{align*}, a number that is even closer to the actual value of \begin{align*} \sqrt{5}\end{align*}. Using the linear approximation again,

Solving for \begin{align*}x\end{align*} by setting \begin{align*}f(x) = 0\end{align*}, we obtain

\begin{align*}x = x_2 = 2.236111,\end{align*}

which is even a better approximation than \begin{align*}x_1 = 9/4\end{align*}. We could continue this process generating a better approximation to \begin{align*} \sqrt{5}\end{align*}. This is the basic idea of Newton’s Method.

Here is a summary of Newton’s method.

Newton’s Method

Guess the first approximation to a solution of the equation \begin{align*}f(x) = 0\end{align*}. A graph would be very helpful in finding the first approximation (see figure below).

Use the first approximation to find the second, the second to find the third and so on by using the recursion relation

To help us find the first approximation, we make a graph of \begin{align*}f(x)\end{align*}. As Figure 11 suggests, set \begin{align*}x_1 = 0.6\end{align*}. Then using the recursion relation, we can generate \begin{align*}x_2\end{align*}, \begin{align*}x_3, \ldots\end{align*}.

Use the linearization method to show that when \begin{align*}x \ll 1\end{align*} (much less than \begin{align*}1\end{align*}), then \begin{align*}(1 + x)^n \approx 1 + nx\end{align*}. Hint: Let \begin{align*}x = 0.\end{align*}

Use the result of problem #3, \begin{align*}(1 + x)^n \approx 1 + nx\end{align*}, to find the approximation for the following:

\begin{align*}f(x) = (1 - x)^4\end{align*}

\begin{align*} f(x) = \sqrt{1 - x}\end{align*}

\begin{align*} f(x) = \frac{5} {\sqrt{1 + x}}\end{align*}

Without using a calculator, approximate \begin{align*}(1.003)^{99}\end{align*}.

Use Newton’s Method to find the roots of \begin{align*}x^3 + 3 = 0\end{align*}.